Verified Path Weight — Problem Statement & Solution Guide
Problem Description
You are given an integer array nums. Compute and return the arithmetic sum of all elements contained in nums. The function should process the entire array in linear time and use only O(1) additional memory beyond the input storage.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Verified Path Weight"
WHY DOES IT MATTER?
The single-pass accumulation pattern is essential because it guarantees linear time and constant space, which are the most efficient bounds for this problem. It eliminates the need for auxiliary data structures or multiple traversals, making the algorithm both simple and scalable.
OPTIMIZATION CHALLENGE
The key insight is that the sum of a sequence can be built incrementally: each new element only needs to be added to the current total. This eliminates the need for nested loops or temporary arrays, reducing both time and space complexity.
REAL-WORLD CONNECTION
In real-time financial trading platforms, calculating the total volume of trades in a minute requires processing millions of records quickly. A single-pass sum allows the system to update the total in real time without storing all trade amounts, ensuring low latency and high throughput.
When explaining this pattern in an interview, emphasize the importance of avoiding unnecessary data structures. Show that you can achieve the goal with a simple loop and an accumulator, and discuss how this pattern generalizes to other problems like computing prefix sums or sliding window aggregates.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem of computing the arithmetic sum of an array is a classic example of a linear-time, constant-space algorithm. A naive approach might involve nested loops or repeated scans, leading to O(n^2) time or unnecessary memory usage. However, the optimal solution leverages a single pass through the array, maintaining an accumulator that aggregates the sum as each element is processed. This pattern—often called the "single-pass accumulation"—ensures that the algorithm runs in O(n) time while using only O(1) additional memory, regardless of input size.
In large-scale systems, such as real-time analytics or streaming data pipelines, the ability to process data in a single pass is critical. It reduces CPU cycles, minimizes cache misses, and eliminates the need for intermediate storage that could become a bottleneck. Moreover, constant space guarantees that the algorithm scales to arrays that may not fit entirely in memory, allowing for efficient streaming or chunked processing.
The underlying theory is rooted in the principle of linearity of summation: the sum of a sequence can be expressed as the sum of its prefixes. By iteratively adding each element to a running total, we avoid recomputation and achieve optimal performance. This paradigm is a foundational building block for many higher-level algorithms, such as prefix sums, cumulative distribution functions, and sliding window computations.
Interview Questions on This Problem
Q1What is the time and space complexity of summing all elements in an integer array, and why is this considered optimal?
The optimal solution runs in O(n) time, where n is the number of elements, because each element is visited exactly once. It uses O(1) additional space, as only a single accumulator variable is needed. This is optimal because any algorithm must inspect each element at least once to compute the sum, and no extra storage beyond the accumulator is required.
Q2How would you modify the algorithm if the array could contain very large integers that might overflow a 32-bit integer?
Use a larger numeric type such as 64-bit long (long in Java, long long in C++, or BigInteger in Java for arbitrary precision). The algorithm remains the same—iterate and accumulate—but the accumulator type must accommodate the larger range to prevent overflow.
Q3In a distributed system where the array is partitioned across multiple nodes, how can you compute the global sum efficiently?
Each node computes a local sum of its partition in O(k) time (k elements on that node). Then, a reduction operation (e.g., tree-based or map-reduce) aggregates the local sums into a global sum. This parallel approach maintains linear total work and only requires O(1) additional space per node beyond the local accumulator.
Examples
Input
[3, -2, 7, 0]
Output
8
Explanation: Add each element sequentially: 3 + (-2) = 1, 1 + 7 = 8, 8 + 0 = 8. The final sum is 8.
Input
[5]
Output
5
Explanation: The array contains a single element, so the sum equals that element: 5.
Input
[-10, 20, -5, 15]
Output
20
Explanation: Compute step‑by‑step: -10 + 20 = 10, 10 + (-5) = 5, 5 + 15 = 20. The resulting sum is 20.
Constraints
- 1 <= nums.length <= 100000
- -1000000000 <= nums[i] <= 1000000000
- The result fits within a signed 64‑bit integer.
Optimal Approach & Strategy
The optimal approach initializes a single accumulator to zero, iterates through the array once, adds each element to the accumulator, and returns the accumulator. This runs in O(n) time and uses O(1) additional space.
Brute Force Approach
A naive approach might use a nested loop to repeatedly sum subsets of the array, or create a copy of the array and repeatedly pop elements, leading to O(n^2) time or O(n) extra space. This is inefficient for large inputs.
Verified Code Solutions
function solution(nums) {
// JavaScript solution
return nums.reduce((a, b) => a + b, 0);
}class Solution {
public:
int solution(vector<int> nums) {
// C++ solution
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
// Java solution
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
}def solution(nums):
# Python solution
return sum(nums)function solution(nums) {
// JavaScript solution
return nums.reduce((a, b) => a + b, 0);
}Asked in Top Tech Interviews
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